Results 1 to 7 of 7

Math Help - Finding derivatives

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    5

    Finding derivatives

    y= (x-2)/(x^2-x+1)

    Find y` and y``

    for the problem i get to
    y`=(-x^2+4x-1)/(x^2-x+1)^2 by u/v method but the answer of this derivative problem is (Given in the book)
    -3(x^2-3x+1)/(x^2-x+1)^3
    How can i Get to this form ?
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    y=\frac{x-2}{x^2-x+1}

    I used the quotient rule (I think you did too):

    \frac{dy}{dx}=\frac{v\frac{du}{dx}-u \frac{dv}{dx}}{v^2}

    where u is the numerator and v is the denominator. In this case:

    u=x-2 and v=x^2-x+1.

    So I did it like this:

    \frac{dy}{dx}=\frac{(x^2-x+1)(1)-(x-2)(2x-1)}{(x^2-x+1)^2}

    \frac{dy}{dx}=\frac{x^2-x+1-2x^2+5x-2}{(x^2-x+1)^2}

    \frac{dy}{dx}=\frac{-x^2+4x-1}{(x^2-x+1)^2}

    So you're right!

    I think it's a typo or just this in another form.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    Posts
    5
    Quote Originally Posted by Showcase_22 View Post

    I think it's a typo or just this in another form.
    Yeah! but i have to come in that form somehow and i want to know the way.Besides, i have to go for second derivative of this function too
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    If you put x=1 into both the equations you get two different answers so they're not the same.

    The only way to get (x^2-x+1)^3 as the denominator is to differentiate again. I thought this would be f''(x).
    However, I began to differentiate again (using the quotient rule, it's similar to before) and I got a cubic appearing on top (hence this can't be f''(x)).

    Try and differentiate f'(x) again and see what you get. I'm pretty sure i'm doing it correctly.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2008
    Posts
    5
    Quote Originally Posted by Showcase_22 View Post
    If you put x=1 into both the equations you get two different answers so they're not the same.

    The only way to get (x^2-x+1)^3 as the denominator is to differentiate again. I thought this would be f''(x).
    However, I began to differentiate again (using the quotient rule, it's similar to before) and I got a cubic appearing on top (hence this can't be f''(x)).

    Try and differentiate f'(x) again and see what you get. I'm pretty sure i'm doing it correctly.
    Answer of f``(x) seems to be
    6x(2x^2-8x+5)/(x^2-x+1)^4 from the book
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Well I think we can safely say the answer in the back of the book for f'(x) is a typo.

    The good news is that you're doing it correctly!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2008
    Posts
    5
    Quote Originally Posted by Showcase_22 View Post
    Well I think we can safely say the answer in the back of the book for f'(x) is a typo.
    Actually i m not sure about it. So, anybody here can try this one and make a comment.This answer is declared correct by my instructor.By the way it is the problem from anton's book.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding derivatives.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2010, 12:38 AM
  2. Finding derivatives with ln
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 6th 2009, 05:24 PM
  3. Finding Derivatives
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 12th 2009, 07:03 AM
  4. Finding derivatives
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 10th 2009, 01:07 PM
  5. Finding derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 23rd 2009, 08:06 AM

Search Tags


/mathhelpforum @mathhelpforum